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%%文档的题目、作者与日期
\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{复变函数测验2}
%\date{\vspace{-3ex}}
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{2024 年 5 月 13 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %Problem 1
考虑初等多值函数 $f(z)=\mathrm{Ln}(z)$. 
计算 $f(3+4i)$ 的所有值。

\vspace{0.2cm}

%{\color{red}解答：设 $z=x+yi=3+4i$. 则其模长和幅角分别为
%\begin{eqnarray*}
%|z| &=& \sqrt{x^2+y^2} = \sqrt{3^2+4^2}=5, \\
%\mathrm{arg}(z) &=& \arctan\frac{y}{x}=\arctan\frac{4}{3}. 
%\end{eqnarray*}
%根据对数函数的计算公式 $\mathrm{Ln} z = \ln|z| + i\mathrm{arg}(z) + 2k\pi$, 可得
%\begin{eqnarray*}
%\mathrm{Ln}(3+4i) = \ln 5 + i\arctan\frac{4}{3} + 2k\pi, \,\, k=0,\pm 1, \pm 2, \cdots. 
%\end{eqnarray*}
%
%}

\vspace{4cm}

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\item  %Problem 2
考虑初等多值函数 $f(z)=2^z$. 
计算 $f(3+4i)$ 的所有值。

\vspace{0.2cm}

%{\color{red}解答：
%根据一般指数函数的定义，$a^z=e^{z\mathrm{Ln}(a)}$, 因此有 
%\begin{eqnarray*}
%2^{3+4i} &=& e^{(3+4i)\mathrm{Ln}(2)} \\ 
%&=& e^{(3+4i) (\ln 2 +2k\pi i)} \\ 
%&=& e^{(3\ln 2 -8k\pi) + i(4\ln 2 + 6k\pi)} \\
%&=& e^{(3\ln 2 -8k\pi)} [\cos(4\ln 2 + 6k\pi ) + i\sin( 4\ln 2 + 6k\pi )] \\ 
%&=& e^{(3\ln 2 -8k\pi)} [\cos(4\ln 2) + i\sin( 4\ln 2 )], \,\, k=0,\pm 1, \pm 2, \cdots. 
%\end{eqnarray*}
%
%}

\vspace{4cm}

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\item  %Problem 3
考虑初等多值函数 $f(z)=\sqrt[4]{z}$. 
计算 $f(\sqrt{3}+i)$ 的所有值。

\vspace{0.2cm}

%{\color{red}解答：
%根据根式函数的计算公式，$\sqrt[n]{z} = \sqrt[n]{ re^{i\theta} } = \sqrt[n]{r} e^{i\frac{\theta+2k\pi}{n}}$, 因此有 
%\begin{eqnarray*}
%\sqrt[4]{\sqrt{3}+i} &=& \sqrt[4]{ 2e^{i\frac{\pi}{6}} } \\ 
%&=& \sqrt[4]{ 2} e^{ i\frac{ \frac{\pi}{6}+2k\pi}{4} } \,\, (k=0,1,2,3) \\
%&=& \sqrt[4]{ 2}e^{i\frac{\pi}{24}},\,\, \sqrt[4]{ 2}e^{i\frac{13\pi}{24}}, \,\, \sqrt[4]{ 2}e^{i\frac{25\pi}{24}}, \,\, \sqrt[4]{ 2}e^{i\frac{37\pi}{24}}. 
%\end{eqnarray*}
%
%}

\vspace{4cm}

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\item  %Problem 4
设解析函数 $f(z)=\sqrt[4]{z}$ 定义在 $\mathbb{C}-(-\infty,0]$ 上，并且 $f(1)=i$, 试求 $f(1+i)$ 的值。

\vspace{0.2cm}

%{\color{red}解答：函数 $f(z)$ 的自变量 $z$ 的定义域为去掉负实轴的复平面，即 
%\begin{eqnarray*}
%z=r(z)e^{i\theta(z)},\,\, r(z)>0,\,\, -\pi< \theta(z) < \pi, 
%\end{eqnarray*}
%其中 $r(z)$ 和 $\theta(z)$ 分别表示 $z$ 的模长和幅角。函数 $f(z)$ 的所有单值分支为 
%\begin{eqnarray*}
%f_k(z) = \sqrt[4]{r(z)} e^{i \frac{\theta(z)+2k\pi}{4} },\,\, k=0,1,2,3. 
%\end{eqnarray*}
%由题设条件 $f(1)=i$,  因为 $r(1)=1, \theta(1)=0$ 以及 $i=e^{i\frac{0+2\pi}{4}}$, 所以 $k=1$. 所以 
%\begin{eqnarray*}
%f(1+i)=f_1(1+i) = \sqrt[4]{r(1+i)} e^{i \frac{\theta(1+i)+2\pi}{4} } 
%= \sqrt[8]{2} e^{i \frac{\frac{\pi}{4}+2\pi}{4} } =\sqrt[8]{2}e^{i\frac{9\pi}{16}}. 
%\end{eqnarray*}
%}

\vspace{0.2cm}
\newpage

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\item  %Problem 5
设解析函数 $h(z)=\mathrm{Ln}\,(z^2-4)$ 
定义在 $G=\mathbb{C}-(-\infty,2]$ 上，并且 $h(3)=\ln(5)+2\pi i$, 试求 $h(2i)$ 的值。
%在割去从 $-1$ 到 $i$ 的直线段和从 $i$ 到1的直线段、与射线 $x=0$且 $y\ge 1$ 的 $z$ 平面内能分出单值解析分支。并求 $z=0$ 时函数值等于零的那一支在 $z=2$ 的值。

%\vspace{0.2cm}

%{\color{red}解答：多值解析函数 $\mathrm{Ln}\,(z^2-4)$ 的支点为 $z=\pm 2$ 与 $\infty$. 
%在定义域 $G$ 里没有围绕任意一个支点的闭路，所以 $h(z)$ 在 $G$ 中可以确定单值解析分支。
%根据题设条件可得本题的单值分支。
%\begin{eqnarray*}
%h(z) = \mathrm{Ln}(z^2-4) &=& \ln |z^2-4| + i \mathrm{Arg} (z^2-4), \\ 
%h(3) &=& \ln(5)+i \mathrm{arg} (5) + 2\pi i. 
%%\Rightarrow h(z) &=& \ln |z^2-4| + i \mathrm{Arg} (z^2-4). 
%\end{eqnarray*}
%在 $G$ 中取一条从 $z_1=3$ 到 $z_2=2i$ 的路径，记为 $C$. 于是有 
%\begin{eqnarray*}
%h(z_2) &=& \ln |z_2^2-4| + i \mathrm{Arg} (z_2^2-4) \\
%&=& \ln(8) + i \mathrm{Arg} (z_1^2-4) + \Delta_C \mathrm{Arg} (z^2-4) \\
%&=& \ln(8) +i \mathrm{arg} (5) + 2\pi i + \Delta_C \mathrm{Arg} (z+2) + \Delta_C \mathrm{Arg} (z-2) \\
%&=& \ln(8) + 2\pi i + i\arctan\frac{2}{2} + i(\pi - \arctan\frac{2}{2}) \\
%&=& \ln(8) + 3\pi i. 
%\end{eqnarray*}
%
%}

\vspace{6cm}

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\item  %Problem 6
设 $f(z)=\frac{z}{1-z}$, 证明当 $|z|<1$ 时，有
$\mathrm{Re} \left[ 1+ z\frac{f\,''(z)}{f\,'(z)} \right] >0. $

%\vspace{0.2cm}

%{\color{red}解答： 直接计算导数，可得 $f'(z) = \frac{1}{(1-z)^2}$ 和 $f''(z) = \frac{2}{(1-z)^3}$. 代入所证表达式，可得 
%\begin{eqnarray*}
%1+ z\frac{f\,''(z)}{f\,'(z)} = 1+ z\frac{2/(1-z)^3}{1/(1-z)^2} = \frac{1+z}{1-z} = \frac{(1+z)(1-\bar{z})}{(1-z)(1-\bar{z})}. 
%\end{eqnarray*}
%设 $z=x+iy$, 则上式等于
%\begin{eqnarray*}
%\frac{(1+x+iy)(1-x+iy)}{(1-x-iy)(1-x+iy)} = \frac{(1-x^2-y^2) + 2yi}{(1-x)^2+y^2}. 
%\end{eqnarray*}
%因为 $|z|<1$, 所以 $1-x^2-y^2>0$, 所以上式的实部为正。
%}

\vspace{6cm}

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\item  %Problem 7
设函数 $f(z)$ 如下，证明 $f(z)$ 在原点满足柯西-黎曼方程，但是不可微。
$$f(z) = \left\{ \begin{array}{ll}
\frac{x^3-y^3+i(x^3+y^3)}{x^2+y^2}, & z=x+iy\neq 0, \\
0, & z=0,
\end{array}\right. $$

%\vspace{0.2cm}

%{\color{red}解答：
%首先验证柯西-黎曼条件。现有 $f(z)=u(x,y)+iv(x,y)$, 其中
%\begin{eqnarray*}
%u(x,y)= \left\{ \begin{array}{ll} \frac{x^3-y^3}{x^2+y^2}, & (x,y)\neq (0,0), \\ 0, & (x,y)=(0,0), \end{array}\right.
%%\,\,\mathrm{与}\,\,
%v(x,y)= \left\{ \begin{array}{ll} \frac{x^3+y^3}{x^2+y^2}, & (x,y)\neq (0,0), \\ 0, & (x,y)=(0,0).  \end{array}\right.
%\end{eqnarray*}
%要验证 $u_x(0,0)=v_y(0,0)$ 与 $u_y(0,0)=-v_x(0,0)$. 
%按照偏导数的定义，可得
%\begin{eqnarray*}
%u'_x(0,0) &=& \lim\limits_{x\to 0}\frac{u(x,0)-u(0,0)}{x} = \lim\limits_{x\to 0}\frac{x-0}{x} = 1, \\ 
%u'_y(0,0) &=& \lim\limits_{y\to 0}\frac{u(0,y)-u(0,0)}{y} = \lim\limits_{y\to 0}\frac{-y-0}{y} = -1, \\ 
%v'_x(0,0) &=& \lim\limits_{x\to 0}\frac{v(x,0)-v(0,0)}{x} = \lim\limits_{x\to 0}\frac{x-0}{x} = 1, \\ 
%v'_y(0,0) &=& \lim\limits_{x\to 0}\frac{v(0,y)-v(0,0)}{y} = \lim\limits_{y\to 0}\frac{y-0}{y} = 1. 
%\end{eqnarray*}
%因此柯西-黎曼条件 $u'_x=v'_y, u'_y=-v'_x$ 在原点成立。 
%根据可微的定义，要计算
%\begin{eqnarray*}
%f'(0,0) = \lim\limits_{(x,y)\to (0,0)} \frac{x^3-y^3+i(x^3+y^3)}{(x^2+y^2)(x+iy)} 
%\overset{y=kx}{=} \lim\limits_{x\to 0} \frac{[(1-k^3)+(1+k^3)i]x^3}{(1+k^2)(1+ki)x^3} 
%= \frac{(1-k^3)+(1+k^3)i}{(1+k^2)+(k+k^3)i}. 
%\end{eqnarray*}
%这是一个与 $k$ 有关的表达式，因此 $f'(0,0)$ 不存在，即 $f(z)$ 在原点不可微。
%
%}

%\vspace{0.2cm}


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\end{enumerate}


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\end{document}

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